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Suppose that $\alpha>0$. Then $\mathbb{Q}$ has Hausdorff dimension $0$, so for each $n$ there are intervals $(a_{k},b_{k})$ such that $\mathbb{Q}\cap[0,1]\subseteq\bigcup_{k}(a_{k},b_{k})$ and $\sum_{k}(b_{k}-a_{k})^{\alpha}<\frac{1}{n}$. Let $E_{n,\alpha}=\bigcup_{k}(a_{k},b_{k})$. Then each $E_{n,\alpha}$ is a dense open set, and hence each $E_{n,\alpha}$ has Hausdorff dimension $1$ (and more). However, $\bigcap_{n}E_{n,\alpha}$ has $\alpha$ Hausdorff measure $0$. In particular, $\bigcap_{n,m}E_{n,\frac{1}{m}}$ has Hausdorff dimension zero. In fact, generalizing this example, we can show that for each set $E$ of Hausdorff dimension $0$ there is a dense $G_{\delta}$ set $G$ with $E\subseteq G$ where $G$ has Hausdorff dimension $0$ as well.

I also have a counterexample from potential theory. Let $\{x_{k}|k\geq 1\}$ be a countable dense subset of $[0,1]$ and let $\sum_{k=1}^{\infty}a_{k}<\infty$. Let $f(z)=\sum_{k=1}^{\infty}a_{k}\mathrm{Log}(x-x_{k})$, and let $E=f^{-1}[\{-\infty\}]$. Let $E_{n}=f^{-1}[\infty,-n)\cap[0,1]$. Then each $E_{n}$ is a dense open subset of $[0,1]$. However, $E=\bigcap_{n}E_{n}$ has Hausdorff dimension zero. In fact, in potential theory we say that $E$ is a polar set.